# Electromagnetics: find D,E,P between two parallel conducting plates

• larginal
In summary, the conversation discusses a problem in electromagnetics homework and the use of Gauss' law in solving it. The speaker shares their attempt at solving the problem and asks for feedback. They also provide advice about using the correct field and accounting for differences in the dielectric. Overall, the speaker believes the approach to the problem is correct.
larginal
Homework Statement
Two infinite, parallel, perfectly conducting plates of a negligible thickness intersect the z-axis at right angles at z= 0.2 m and z = 0, carrying uniform charge densities p(surface charge density) = 45 * 10^-6 and -45*10^-6 respectively. the lower half of the space(0 <z<1) is filled with a dielectric of epsilon r = 1.5 , and the upper half of the space (0.1 < z < 0.2) is the air, epsilon r = 1.
when the lower plate is maintained at potential V = 0,
find
(a) D, E, P and V in the region 0 < z < 0.2 m
(b) total surface charge density on each plate.
Relevant Equations
D = epsilon0 * E + P
I tried to solve this problem on my own, but I'm not sure whether I solved correctly or not.
it is electromagnetics homework from my Uni, and it is pretty tough for me.
I attached the image of the problem and how I tried to solve this one.
I hope somebody will give some feedback.

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If possible, please type in your work rather than post photos of your work. It makes it easier to read and easier to quote specific parts of your work.

In using Gauss' law in the form ##\oint \mathbf{D} \cdot \mathbf{dS} = \large \frac{Q_{\rm enclosed}^{\rm free}}{\varepsilon_0}##, the ## \mathbf{D}##-field appearing in the integral is the field due to all of the charge in the system. So, when you used the blue Gaussian surface to find ##\mathbf{D}##, you found the net field due to all of the charge. When you used the red Gaussian surface, you just found the same field by a different choice of Gaussian surface. You should not add the ##\mathbf{D}##'s for the two methods.

When you integrated ##\mathbf{E} = -\boldsymbol{\nabla} V ## to find ##V##, you need to take into account that ##\mathbf{E}## in the air is different than ##\mathbf{E}## in the dielectric.

Otherwise, I think you have the correct approach to the problem.

## 1. What is the purpose of finding D, E, and P between two parallel conducting plates in electromagnetics?

The purpose of finding D, E, and P between two parallel conducting plates is to understand the behavior of electric fields and charges in this specific configuration. This can help in designing and analyzing various devices such as capacitors, transmission lines, and parallel plate antennas.

## 2. How do you calculate the electric displacement (D) between two parallel conducting plates?

The electric displacement (D) between two parallel conducting plates can be calculated by dividing the magnitude of the charge on one plate by the distance between the plates. This can also be expressed as the product of the permittivity of the medium between the plates and the electric field strength.

## 3. What is the relationship between electric field (E) and electric displacement (D) between two parallel conducting plates?

The relationship between electric field (E) and electric displacement (D) between two parallel conducting plates is that D is proportional to E. This means that as the electric field strength increases, the electric displacement also increases.

## 4. How do you find the polarization (P) between two parallel conducting plates?

The polarization (P) between two parallel conducting plates can be found by subtracting the electric displacement (D) from the electric field strength (E). This represents the contribution of the bound charges on the plates to the overall electric field.

## 5. What factors can affect the values of D, E, and P between two parallel conducting plates?

The values of D, E, and P between two parallel conducting plates can be affected by factors such as the distance between the plates, the permittivity of the medium between the plates, and the magnitude of the charges on the plates. Additionally, the presence of dielectric materials between the plates can also impact these values.

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